Molarity calculations are a “must learn” in general or organic labs. Before going into Molarity calculations, it is important that you brush up on your stoichiometry. Chemistry Molarity Calculations
We must also understand what concentration is. I like the definition which my chemistry book gives.
It states “Concentration is a measure of the quantity of solute dissolved in a given quantity of solutions“.
This definition was taken from Silberberg’s Fourth Edition Chemistry book – Chemistry, The Molecular Nature of Matter and Change. To sum up, what this means, this means that if you add more of the solute in the solution, you will have a higher concentration. If this doesn’t make sense right now, don’t worry!
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Now let’s talk about what Molarity is. Molarity is just another unit, abbreviated M (not to be mistaken for moles). The unit for Molarity is Moles of solute per Liter of the solution or (mol/L).
Let’s dig into the main concept in this chapter… How to solve Molarity calculations.
In most all Molarity problems, you will be given three out of four variables and will solve for the fourth. The four variables are:
The concentration of A
The volume of A
The concentration of B
The volume of B
A general formula for solving Molarity problems is the following:
Mdil x Vdil = number of moles = Mconc x Vconc
Don’t think you need to solve every Molarity problem using the above equation, you can also do them by unit conversions. Chemistry Molarity Calculations
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Let me do a sample Molarity problem so you can see how the above equation works.
|Problem: If 5.0 L of 3.50 M acid is to makes a 0.500 M solution, what is the final volume of the solution when diluted to 0.500 M? Assume this is a 1 to 1 mole ratio.
Plan: In order to solve this problem, we will use the above equation Mdil x Vdil = number of moles = Mconc x Vconc. We know we have the concentration (molarity) of A, the volume of A and the concentration of B. Therefore, let’s solve for the volume of B.
(Mdil x Vdil)/Mconc = Vconc
Answer: By filling in the above algebra equation we get:
Discussion: A good way to determine if you are performing the problem correctly, is to check if your units cancel to give what the problem is asking for.
Here both the molarities cancel out and give liters. Because the problem states that this was a 1 to 1 mole ratio,
the “=number of moles” can be ignored. This is the case in about 60% of Molarity problems.
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